grep/awk 3rd column containing string

Question

Example of string

blah:blah:example - 1234
blah1:blah2:example2 - 3213

What I want to check exists

- 1234

but numeric can be different

I know I can use

awk -F: '3ドル ~ / - /' 

to check if " - " exists, but how can I apply numeric possibilities?

Answer 1

So you basically want to check if after ' - ' there's some numbers?
If you want to do this you can write
awk -F: '3ドル ~ / - [0-9]+/'

More advanced solution

If you have an array of possible values for your last number (es. 1, 2, 3, 4, 5) you could write something like this.

values=(1 2 3 4 5)
awk -v vals="$(echo ${values[@]})" \
'{if(match(vals,3ドル" ")){print 0ドル}}' input_file.txt

This will print only those entries of input_file.txt having the last number in the values array you specified.

Answer 2

The following awk command will display all lines from file that has a - in its penultimate whitespace-delimited column, and a positive integer in its last whitespace-delimited column:

awk '$(NF-1) == "-" && $NF ~ /^[0-9]+$/' file

The following grep command would do the same thing:

grep -E -e '-[[:space:]]+[0-9]+$' file

The -E is needed to understand the + modifier in the pattern, and -e is needed to stop grep from interpreting the - in the pattern as a command line option (-e could also be changed replaced by --, which signals the end of any command line options).